Decide wether $x^6 + 2x^5 - 5x^4 - 5x^3 - 5x^2 + 2x + 1 = 0$ can be solved by radicals

99 Views Asked by At

Decide wether a given equation (over $\mathbb{Q}$) can be solved by radicals and, if it is possible, find a radical sequence to solve it.

The main problem is that I'm not sure what they want from me, because $x^6 + 2x^5 - 5x^4 - 5x^3 - 5x^2 + 2x + 1$ is a reciprocal palindromic equation, so it can be solved by substituting $x + \frac{1}{x} = y$, or $0 = x^6 + 2x^5 - 5x^4 - 5x^3 - 5x^2 + 2x + 1 = (x^2 + x + 1)(x^4 + x^3 - 7x^2 + x + 1)$ and since their degrees are less than $5$, they can be solved by radicals, hence I guess original equation can be solved by radicals. Is this the right approach? And what after that? How do I find its radical sequence and solve the equation by using that sequence?